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B Are differential angles vectors?

  1. Mar 11, 2016 #1
    Because on the book it is said that little angles are vectors but my question is:
    Are they vectors at all or they are scalar and we assign them a direction by multiplying them by a versor? The same for angular velocity, is it a vector at all or we made it a vector for making the right hand rule work?
     
  2. jcsd
  3. Mar 11, 2016 #2
    A scalar has magnitude only, such as temperatures at different points in a room.

    A vector has not only magnitude but direction, say the flow of water during the draining of a bath tub.

    So if I want to give you directions from you to me, saying I am five miles away [magnitude] is not enough; I need to tell you 'fives miles away to the west' , for example, using a compass as the scale.

    https://en.wikipedia.org/wiki/Euclidean_vector
     
  4. Mar 11, 2016 #3

    finite angular displacements are not vectors as they do not obey the the rule that A + B = B+ A ,
    but in infinitesimal rotations the displacements can have a direction along the axis of rotations - clockwise/anticlockwise rotation can have two directions
    and the time rate of change of angles do provide a vector called angular velocity- no doubt the rotations in general can have three components like the normal vectors or generalized rotations......
     
  5. Mar 11, 2016 #4
    unlikely. best to give actual quotes so we know what your reference is.

    'Little angles' are angles, not vectors, right? For one thing an angle is usually not referenced to a coordinate frame of reference, like a graph plot. Vectors have direction because they have a specific direction with respect to a frame.
     
  6. Mar 13, 2016 #5

    vanhees71

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    2016 Award

    I don't know, what you mean by "little vectors", but an infinitesimal rotation can be written in terms of an axial vector ##\delta \vec{\varphi}##:
    $$\delta \vec{V}=\delta \vec{\varphi} \times \vec{V}.$$
    To see this look at a rotation around an axis ##\vec{n}## (right-hand rule!) with an angle ##\phi##. If you take the ##z## axis of a Cartesian righthanded coordinate system then it's described by the matrix
    $$\hat{D}=\hat{D}_{\vec{n}}(\varphi)=\begin{pmatrix} \cos \varphi & -\sin \varphi &0 \\
    \sin \varphi & \cos \varphi & 0 \\
    0 & 0 & 1
    \end{pmatrix},$$
    i.e.,
    $$\hat{D} \vec{V}=V_z \vec{e}_z + \cos \varphi (V_x \vec{e}_x + V_y \vec{e}_y)+\sin \varphi (-V_y \vec{e}_x+V_x \vec{e}_y).$$
    On the other hand we have
    $$\vec{e}_z \times \vec{V}=\vec{n} \times \vec{V}=-V_y \vec{e}_x+ V_x \vec{e}_y, \quad \vec{n} \times (\vec{n} \times \vec{V}) = -V_x \vec{e}_x-V_y \vec{e}_y,$$
    from which we find
    $$\hat{D} \vec{V}=\vec{n} (\vec{n} \cdot \vec{V})-\cos \varphi \vec{n} \times (\vec{n} \times \vec{V}) + \sin \varphi \vec{n} \times \vec{V}.$$
    For a small angle ##\delta \varphi## this implies
    $$\hat{D} \vec{V}=\vec{n} (\vec{n} \cdot \vec{V}) - \vec{n} \times (\vec{n} \times \vec{V}) +\delta \varphi \vec{n} \times \vec{V} + \mathcal{O}(\delta \varphi^2).$$
    Now we have
    $$\vec{n} \times (\vec{n} \times \vec{V}) =\vec{n} (\vec{n} \cdot \vec{V})-\vec{V} (\vec{n} \cdot \vec{n}) = \vec{n} (\vec{n} \cdot \vec{V})-\vec{V}$$
    and thus finally
    $$\hat{D} \vec{V}=\vec{V} +\delta \varphi \vec{n} \times \vec{V} + \mathcal{O}(\delta \varphi^2).$$
     
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